Sign In
LHS-2≥RHS-2
LHS-x≥RHS-x
.LHS /2.≥.RHS /2.
The solution set to the inequality is x≥ -4. Therefore, we should shade everything that is greater than or equal to -4. Note that this is a non-strict inequality which means we will keep the endpoint closed.
Inequality:& 2x^2-5x<12
Equality:& 2x^2-5x=12
Let's solve it by using the Quadratic Formula.
LHS-12=RHS-12
Use the Quadratic Formula: a = 2, b= -5, c= -12
- (- a)=a
Calculate power and product
Add terms
Calculate root
State solutions
(I), (II): Add and subtract terms
Calculate quotient
The boundaries of the inequality is x=-1.5 and =4. Let's mark these solutions on a number line. We will keep the points open since this is a strict inequality. To identify where we should shade the number line, we will also include a few test points.
By substituting x with -3, 0, and 5 in the inequality we can determine where the inequality holds true. |c|c|c| [-0.8em] x & 2x^2-5x<12 & Evaluate [0.5em] [-1em] -3 & 2( -3)^2-5( -3)? <12 & 33 ≮ 12 * [0.5em] [-1em] 0 & 2( 0)^2-5( 0)? <12 & 0 < 12 ✓ [0.5em] [-1em] 5 & 2( 5)^2-5( 5)? <12 & 25 ≮ 12 * [0.5em] As we can see, the inequality is true for values between the boundary points. Therefore, we should shade this region.